In the adjoining figure, CM = 5 cm, RB = 9 cm, `CD bot AB`, O is the centre of larger circle and K is the centre of smaller circle. Find the area of shaded region

Explanation

Let radius of larger circle be R and the radius of smaller circle is r.
Since, `CM = 5 cm`
`:. MO = R - 5`
Also `OB = R and RB = 9`
`:. OR = R - 9`
In `Delta AOM`,
`angle1 + angle2 = 90^(@)`...(1)
In `Delta AMR`,
`angle 2 + angle3 = 90^(@)` (angle in a semicircle is right angle) ..(2)
From eqs. (1) and (2), we get
`angle1 + angle 2 = angle + angle 3`
`rArr angle 1 = angle3`
In `Delta s AMO and MOR`
`angle 1 = angle3` (just proved)
`angle 4 = angle 5` (each `90^(@)`)
`:. Delta AMO ~ Delta MRO` (AA corollary)
`:. (MO)/(RO) = (AO)/(MO)`
`rArr (R - 5)/(R - 9) = (R)/(R- 5)`
`rArr (R - 5)^(2) = R (R - 9) rArr R^(2) + 25 - 10R = R^(2) - 9R`
`rArr R = 25 cm`
Now, `2R - 9 = 2r rArr 2r = 41`
`rArr r = 20.5 cm`
`:.` Required area `= pi R^(2) - pi r^(2) = pi (R + r) (R - r)`
`= (22)/(7) xx 45.5 xx 4.5 = 643.5 cm^(2)`